Respuesta :
Answer : The final pressure in the system is 4.22 atm.
Explanation :
First we have to calculate the moles of methane.
[tex]PV=n_1RT[/tex]
where,
P = pressure of gas = 9.61 atm
V = volume of gas = 3.06 L
T = temperature of gas = T
[tex]n_1[/tex] = number of moles of methane gas = ?
R = gas constant
Now put all the given values in the ideal gas equation, we get:
[tex](9.61atm)\times (3.06L)=n_1\times RT[/tex]
[tex]n_1=\frac{29.4}{RT}[/tex]
Now we have to calculate the moles of oxygen gas.
[tex]PV=n_2RT[/tex]
where,
P = pressure of gas = 1.75 atm
V = volume of gas = 6.65 L
T = temperature of gas = T
[tex]n_2[/tex] = number of moles of oxygen gas = ?
R = gas constant
Now put all the given values in the ideal gas equation, we get:
[tex](1.75atm)\times (6.65L)=n_2\times RT[/tex]
[tex]n_2=\frac{11.6}{RT}[/tex]
Now we have to determine the final pressure in the system after mixing the gases.
[tex]P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}[/tex]
where,
[tex]P_{total}[/tex] = final pressure of gas = ?
[tex]V_{total}[/tex] = final volume of gas = (3.06 + 6.65)L = 9.71 L
T = temperature of gas = T
R = gas constant
Now put all the given values in the ideal gas equation, we get:
[tex]P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}[/tex]
[tex]P_{total}=4.22atm[/tex]
Therefore, the final pressure in the system is 4.22 atm.
